与实变函数相同，此处省略

($\Rightarrow$)
令 $f'(z_0) = \displaystyle \lim_{w \to 0} \frac{f(z_0+w) - f(z_0)}{w}, z_0 := x_0 + iy_0$，等价于

$$\lim_{w \to 0} \frac{|f(z_0+w) - f(z_0) - f'(z_0)w|}{|w|} = 0$$

令 $S := f(z_0+w) - f(z_0) - f'(z_0)w$，取

$$\begin{cases} h := \mathrm{Re}(w) \\ k := \mathrm{Im}(w) \end{cases},\quad \begin{cases} \alpha := \mathrm{Re} \circ f' \\ \beta := \mathrm{Im} \circ f' \end{cases}$$

则分子可写为

$$\begin{aligned} S &= f(z_0+w) - f(z_0) - f'(z_0)w \\ &= \underbrace{u(x_0 + h, y_0 + k) + iv(x_0 + h, y_0 + k))}_{f(z_0+w)} \underbrace{- u(x_0, y_0) - iv(x_0, y_0)}_{-f(z_0)} \underbrace{-(\alpha(x_0, y_0) + i\beta(x_0, y_0))(h + ik)}_{-f'(z_0)w} \\ &= \underbrace{u(x_0 + h, y_0 + k) - u(x_0, y_0) - (\alpha(x_0,y_0)h - \beta(x_0, y_0)k)}_{=: R} + i \underbrace{(v(x_0 + h, y_0 + k) - v(x_0, y_0) - (\beta(x_0,y_0)h + \alpha(x_0, y_0)k))}_{=: I} \end{aligned}$$

即原极限

$$\lim_{w \to 0} \frac{|f(z_0+w) - f(z_0) - f'(z_0)w|}{|w|} = \lim_{(h,k) \to (0,0)} \frac{\sqrt{R^2 + I^2}}{\sqrt{h^2 + k^2}} = 0 \iff \underbrace{\lim_{(h,k) \to (0,0)} \frac{|R|}{\sqrt{h^2 + k^2}} = 0}_{\text{condition} \ 1} \wedge \underbrace{\lim_{(h,k) \to (0,0)} \frac{|I|}{\sqrt{h^2 + k^2}} = 0}_{\text{condition} \ 2}$$

其中，condition 1 等价于 $u$ 在 $(x_0,y_0)$ 处 Frechet 可微，且

$$\frac{\partial u}{\partial x}(x_0, y_0) = \alpha(x_0, y_0),\quad \frac{\partial u}{\partial y}(x_0, y_0) = -\beta(x_0, y_0)$$

condition 2 等价于 $v$ 在 $(x_0,y_0)$ 处 Frechet 可微，且

$$\frac{\partial v}{\partial x}(x_0, y_0) = \beta(x_0, y_0),\quad \frac{\partial v}{\partial y}(x_0, y_0) = \alpha(x_0, y_0)$$

得到 C-R 方程成立

($\Leftarrow$)
由于 $f \circ J = u + iv$，则

$$f(z_0 + w) - f(z_0) = \underbrace{u(x_0 + h, y_0 + k) - u(x_0, y_0)}_{=: R} + i\underbrace{(v(x_0 + h, y_0 + k) - v(x_0, y_0))}_{=:I}$$

取 $R, I$ 的一次近似展开

$$\begin{cases} R \approx R_1 := \frac{\partial u}{\partial x}(x_0, y_0) h + \frac{\partial u}{\partial y}(x_0, y_0) k \\ I \approx I_1 := \frac{\partial v}{\partial x}(x_0, y_0) h + \frac{\partial v}{\partial y}(x_0, y_0) k \end{cases}$$

由 C-R 方程可知 $\frac{\partial u}{\partial y}(x_0, y_0) = -\frac{\partial v}{\partial x}(x_0, y_0), \ \frac{\partial u}{\partial x}(x_0, y_0) = \frac{\partial v}{\partial y}(x_0, y_0)$

$$\begin{aligned} f(z_0 + w) - f(z_0) &\approx R_1 + i I_1 \\ &= \left(\frac{\partial u}{\partial x}(x_0, y_0) + i \frac{\partial v}{\partial x}(x_0, y_0)\right) h + \left(\frac{\partial u}{\partial y}(x_0, y_0) + i \frac{\partial v}{\partial y}(x_0, y_0)\right) k \\ &= \left(\frac{\partial u}{\partial x}(x_0, y_0) + i \frac{\partial v}{\partial x}(x_0, y_0)\right) (h + ik) \\ &\stackrel{C-R}{=} \left(\frac{\partial v}{\partial y}(x_0, y_0) - i \frac{\partial u}{\partial y}(x_0, y_0)\right) (h + ik) \end{aligned}$$

所以

$$\left| \frac{f(z_0 + w) - f(z_0)}{w} - \left(\frac{\partial u}{\partial x}(x_0, y_0) + i \frac{\partial v}{\partial x}(x_0, y_0)\right)\right| \to 0 \quad (w \to 0)$$

即 $f$ 在 $z_0$ 处复可微，且

$$f'(z_0) = \frac{\partial u}{\partial x}(x_0, y_0) + i \frac{\partial v}{\partial x}(x_0, y_0) \left( = \frac{\partial v}{\partial y}(x_0, y_0) - i \frac{\partial u}{\partial y}(x_0, y_0)\right)$$

(1)
由 C-R 方程 $f' = u_x + iv_x = v_y - iu_x$，所以在 $\Omega$ 上

$$f' \equiv 0 \iff u_x \equiv u_y \equiv 0, v_x \equiv y_y \equiv 0 \iff u,v \text{ 为常数}$$

即 $f$ 在 $\Omega$ 上恒为常数
(2)
等价于 $u$ 为常数，即 $u_x \equiv u_y \equiv 0$
由 C-R 方程得

$$\begin{cases} v_x = -u_y = 0 \\ v_y = u_x = 0 \end{cases}$$

所以 $v$ 也为常数
(3)
与 (2) 相同
(4)
令 $|f(z)| = c \geq 0$，若 $c=0$ 则 $f \equiv 0$
取 $c \gt 0$，则

$$c^2 = |f|^2 = u^2 +v^2$$

分别对 $x, y$ 偏微分，通过代换 $u_y = -v_y, v_y = u_x$

$$\begin{cases} 0 = u \cdot u_x + v \cdot v_x \\ 0 = u \cdot u_y + v \cdot v_y \end{cases} \iff \begin{pmatrix} 0 \\ 0 \end{pmatrix} = \begin{pmatrix} u & v \\ v & -u \end{pmatrix} \begin{pmatrix} u_x \\ v_x \end{pmatrix}$$

由于 $u^2 + v^2 = c^2 \gt 0$，所以此处矩阵为正则矩阵，方程仅有自明解 $u_x = v_x = 0$
同样可得 $u_y = v_y = 0$